DC MOTORS

Part III1

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Starting of DC Motor

• A starter is a device to start and accelerate a motor.

• A controller is a device to start the motor, control and reverse the

speed of the DC motor and stop the motor.

• While starting the DC motor, it draws the heavy current which

damages the motor.

• The starter reduces the heavy current and protects the system from

damage.

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Need of Starters for DC Motor

• The dc motor has no back emf.

• At the starting of the motor, the armature current is controlled by the

resistance of the circuit.

• The resistance of the armature is low, and when the full voltage is

applied at the standstill condition of the motor, the armature current

becomes very high which damage the parts of the motor.

• Because of the high armature current, the additional resistance is

placed in the armature circuit at starting.

• The starting resistance of the machine is cut out of the circuit when

the machine gains it speeds.

• The armature current of a motor is given by

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• Thus, Ia depends upon E and Ra, if V is kept constant.

• When the motor is first switched ON, the armature is stationary.

• Hence, the back EMF Eb is also zero.

• The initial starting armature current Ias is given by the equation shown

below.

• Since, the armature resistance of a motor is very small, generally less

than one ohm.

• Therefore, the starting armature current Ias would be very large.

• For example – if a motor with the armature resistance of 0.5 ohms is

connected directly to a 230 V supply, then by putting the values in the

equation (2) we will get.

• This large current would damage the brushes, commutator and

windings.

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• As the motor speed increases, the back EMF increases and the difference (V

– E) go on decreasing.

• This results in a gradual decrease of armature current until the motor attains

its stable speed and the corresponding back EMF.

• Under this condition, the armature current reaches its desired value.

• Thus, it is found that the back EMF helps the armature resistance in limiting

the current through the armature.

• Since at the time of starting the DC Motor, the starting current is very large.

• At the time of starting of all DC Motors, except for very small motors, an

extra resistance must be connected in series with the armature.

• This extra resistance is added so that a safe value of the motor is maintained

and to limit the starting current until the motor has attained its stable speed.

• The series resistance is divided into sections which are cut out one by one,

as the speed of the motor rises and the back EMF builds up.

• The extra resistance is cut out when the speed of the motor builds up to its

normal value.

DC Starters

Diagram

It has three terminals L, F and A

The starter consists of starting resistance divided into several sections and

connected in series with the armature.

The three terminals L, F and A of the starter are connected respectively to the

positive line terminal, shunt field terminal and armature terminal.

The other terminals of the armature and shunt field windings are connected to the

negative terminal of the supply.

Operation To start with, the DC supply is switched on with handle in the

OFF position.

The handle is now moved clockwise to the first stud. As soon as

it comes in contact with the first stud, the shunt field winding is

directly connected across the supply, while the whole starting

resistance is inserted in series with the armature circuit

As the handle is gradually moved over to the final stud, the

starting resistance is cut out of the armature circuit in steps. The

handle is now held magnetically by the no-volt release coil

(NVR) which is energized by shunt field current.

• If the supply voltage is suddenly interrupted the no-volt

release coil (NVC) is demagnetized and the handle goes back

to the OFF position under the pull of the spring.

• If no-volt release coil were not used, then in case of failure

of supply, the handle would remain on the final stud.

• If then supply is restored, the motor will be directly

connected across the supply, resulting in an excessive

armature current.

• If the motor is over-loaded (or a fault occurs), it will draw

excessive current from the supply.

• This current will increase the ampere-turns of the over-load

release coil and pull the armature C, thus short-circuiting the

no-volt release coil.

• The no-volt coil is demagnetized and the handle is pulled to

the OFF position by the spring.

• Thus, the motor is automatically disconnected from the

supply.

Drawbacks of 3 point starter

In a three-point starter, the no-volt release coil is connected in

series with the shunt field circuit so that it carries the shunt field

current.

While exercising speed control through field regulator, the field

current may be weakened to such an extent that the no-volt

release coil may not be able to keep the starter arm in the ON

position.

This drawback is overcome in the four point starter.

4 point starter

only difference between a three-point starter and a four-point

starter is the manner in which no-volt release coil is connected.

However, the working of the two starters is the same.

It may be noted that the three-point starter also provides

protection against an open-field circuit.

This protection is not provided by the four-point starter.

APPLICATIONS OF DC MOTORS

MOTORS.. APPLICATIONS…

DC SHUNT MOTOR

LATHES , FANS, PUMPS DISC AND

BAND SAW DRIVE REQUIRING

MODERATE TORQUES.

DC SERIES MOTOR ELECTRIC TRACTION, HIGH SPEED

TOOLS

DC COMPOUND MOTOR

ROLLING MILLS AND OTHER LOADS

REQUIRING LARGE MOMENTARY

TORQUES.

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Numericals on DC Motor

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Q.1: A DC shunt machine is connected to 250 V supply and has an

armature resistance of 0.12 ohm and resistance of field circuit is

100 ohm. Find the ratio of the speed as a generator to the speed as a

motor, line current in each case is 80 A.

Sol.: Given: V = 250 V, IL = 80 A, Ra = 0.12 ohm, Rsh = 100 ohm

Ish = V/Rsh = 250/100 = 2.5 A

For generator:

Iag = IL + Ish = 80+2.5 = 82.5 A

Eg = V + IagRa = 250 + 82.5 x 0.12 = 259.9 V

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For Motor:

Iam = IL – Ish = 80 – 2.5 = 77.5 A

Em = V – IamRa = 250 – 77.5 x 0.12 = 240.7 V

N α E/Φ

Ng/Nm = (Eg x Φm)/(Em x Φg)

Since field current is same

Φg = Φm

Ng/Nm = Eg/Em = 260/240.7 =1.0798 = 1.08

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Q. 2: The armature resistance of a 200 V shunt motor is 0.4 ohm and no load current is 2 A. When loaded and taken an armature current of 50 A, the speed is 1200 rpm. Find approximately the no load speed.

Sol.: Given: Ra = 0.4 ohm, V = 200 V, N1 = 1200 rpm,

Ia0 = 2A, Ia1 = 50 A

E0 = V – Ia0Ra = 200 – 2 x 0.4 = 199.2 V

E1 = V- Ia1Ra = 200 – 50 x 0.4 = 180 V

N α E/Φ

For shunt motor Φ is constant

N1/N0 = E1/E0 = 180/199.2 = 0.9036

N0= N1/0.9009= 1200/0.9036 = 1328.0212 rpm

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Q. 3:A 250 V shunt motor on no load runs at 1000 rpm and takes 5 A. The total armature and shunt field resistance are 0.2 ohm and 250 ohm respectively. Calculate the speed when loaded and taking a current of 50 A. If the armature reaction weakens by 3%.

Sol.: Given: V = 250 V, N0 = 1000 rpm, IL = 5 A, Ra =0.2 ohm, Rsh = 250 ohm,

Ish = V/Rsh = 250/250 = 1 A

Ia0 = IL – Ish = 5-1 = 4 A

Ia1 = 50 – 1 = 49 A

E0 = V – Ia0Ra = 250 – 4 x 0.2 = 249.2 V

E1 = V – Ia1Ra = 250 – 49 x 0.2 = 240.2 V

Φ1 = 0.97 Φ0

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N α E/Φ

N1/N0 = (E1 x Φ0)/(E0 x Φ1) = 240.2 x Φ0/(249.2 x 0.97 Φ0)

N1/N0 = 240.2/(249.2 x 0.97) = 0.9937

N1 = 1000 x 0.9937 = 993.7 rpm

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Q. 4: A shunt generator delivers 50 kW at 250 V when running at 400 rpm. The armature and field resistance are 0.02 ohm and 50 ohm respectively. Calculate the speed of machine when running as a shunt motor and taking 50 kW input at 250 V. Allowed contact drop is 1 V per brush.

Sol.: Given: P = 50 kW, V = 250 V, Ng = 400 rpm, Ra = 0.02 ohm, Rsh = 50 ohm, Vb = 1 V per brush

IL = P/V = 50,000/250 = 200 A

Ish = V/Rsh = 250/50 =5 A

Iag = IL + Ish = 200+5 = 205 A

Eg = V + IagRa+ Vb = 250 + 205 x 0.02 +2 = 256.1 V

ILm = 200 A

Iam = IL – Ish = 200-5 = 195 A

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Em = V – IamRa –Vb = 250 – 195 x 0.02 – 2 = 244.1 V

N α E [for constant Φ in shunt machine]

Ng/Nm = Eg/Em = 256.1/244.1

Nm = 400 x 0.9531 = 381.2573 rpm

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Q. 5: A 4 pole 250 V wave connected shunt motor gives 10

kW when running at 1000 rpm and drawing Ia and Ish of 60

A and 1 A respectively. It has 560 conductors. Its Ra is 0.2

ohm. Assume a drop of 2 V per brush. Determine

(a) Total Torque, (b) Useful Torque, (c) useful Flux per pole,

(d) Rotational losses (e) Efficiency

Sol.: Given: P = 4, A= 2, V = 250 V, Puseful = 10 kW, N =

1000 rpm, Ia = 60 A, Ish = 1 A, Z = 560, Vb = 2 V

Eb = V – IaRa –Vb = 250 – 60 x 0.2 – 2 = 236 V

(a) Total Torque

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T = (236 x 60 x 60)/(2 x 3.14 x 1000) = 135.3 Nm

(b) Useful torque (Tuseful) = Puseful/w = (P x 60)/(2 x 3.14 x n)

Tuseful = (10000 x 60)/(2 x 3.14 x 1000) = 95.5 Nm

(c) Eb = (PφNZ)/(60A)

Φ = (236 x 60 x 2)/ (4 x 1000 X 560) = 0.0126 Wb

Φ = 12.6 mWb

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Q. 6: A 230 V shunt motor delivers 30 hp at the shaft at 1120

rpm. If the motor has an

efficiency of 87% at this load, determine:

a) The total input power.

b) The line current.

Sol.:

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Q. 7: A separately excited DC motor has the following specifications:

Terminal voltage = 250 V, field voltage = 250 V, armature resistance

= 0.03 Ω, field resistance = 250 Ω. Initially the motor was running

at speed = 1103 rpm while supplied by the rated terminal voltage

and the armature current = 120 A. While supplying constant torque,

what is the speed of the motor if the terminal voltage is reduced to

200 V?

Sol.: Torque is constant and Vf is not changed (the field flux will be

constant)

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Exercise

1. A 4 pole 500 V shunt motor takes 7A on no load, the no

load speed of the motor if it takes 122 A at full load.

Armature resistance is 0.2 Ω, contact drop/brush is 1 V,

Armature reaction weakness the field by 40% on full load.

Find the full load speed of the motor.

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